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anonymous
 one year ago
How to exchange the integral order, i can't solve this problem. ∫∫ xy dxdy, D = {(x,y)  2 <= y <= 4, (y^26)/2 <= x <= y+1}
anonymous
 one year ago
How to exchange the integral order, i can't solve this problem. ∫∫ xy dxdy, D = {(x,y)  2 <= y <= 4, (y^26)/2 <= x <= y+1}

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phi
 one year ago
Best ResponseYou've already chosen the best response.0Here is a plot of the region, showing a "typical" rectangle as we integrate x first.

phi
 one year ago
Best ResponseYou've already chosen the best response.0If we integrate dy first, then we evidently have to break the problem into two regions. from x=3 to 1, and then from x=1 to 5

phi
 one year ago
Best ResponseYou've already chosen the best response.0from the lower limit (in the original problem) \[ x= y^2 6 \] we get \[ y= \pm \sqrt{x+6} \] in the region between x=3 and x=1, evidently the lower limit is \( y= \sqrt{x+6}\) and the upper limit is \( y= \sqrt{x+6} \) and we have \[ \int_{3}^{1} \int_{ \sqrt{x+6}}^{+ \sqrt{x+6}} x y\ dy \ dx \] once we get to x=1 we change the lower limit on y. we originally had x=y+1, from which we get y= x1 the second iterated integral is \[ \int_{1}^{5} \int_{x1}^{+ \sqrt{x+6}} x y\ dy \ dx \]
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